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Inverse of a Matrix by Elementary Operations – Matrices | Class 12 Maths

  • Difficulty Level : Medium
  • Last Updated : 17 Nov, 2020

The Gaussian Elimination method is also known as the row reduction method and it is an algorithm that is used to solve a system of linear equations. It is usually understood as a sequence of operations performed on the corresponding matrix of coefficients. This algorithm is used to find :

  • The rank of a matrix.
  • The determinant of a matrix.
  • The inverse of a matrix.

The operations we can perform on the matrix to modify are:

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  • Interchanging/swapping two rows.
  • Multiplying or Dividing a row by a positive integer.
  • Adding or subtracting a multiple of one row to another.

Now using these operations we can modify a matrix and find its inverse. The steps involved are:



  • Step 1: Create an identity matrix of n x n.
  • Step 2: Perform row or column operations on the original matrix(A) to make it equivalent to the identity matrix.
  • Step 3: Perform similar operations on the identity matrix too. 

Now the resultant identity matrix after all the operations is the inverse matrix.

Examples

Note: 

Here, R1: Row 1, R2: Row 2, R3: Row 3

Example 1: Find the inverse of the following matrix by elementary operations?

A =\begin{bmatrix}2&0&3\\-1&3&-4\\-3&1&-4\end{bmatrix}

Solution:

Let’s perform row or column operations on the original matrix(A) to make it equivalent to the identity matrix.

Step 1: Interchange R2 and R3 rows (to make A[2][2] = 1)



\begin{bmatrix}2&0&3\\-3&1&-4\\-1&3&-4\end{bmatrix}

Step 2: R1 = R1 + R3 (to make A[1][1] = 1)

\begin{bmatrix}1&3&-1\\-3&1&-4\\-1&3&-4\end{bmatrix}

Step 3: R2 = R2 – 3R3 (to make A[2][1] = 0)

\begin{bmatrix}1&3&-1\\0&-8&8\\-1&3&-4\end{bmatrix}

Step 4: R3 = R3 + R1 (to make A[3][1] = 0)

\begin{bmatrix}1&3&-1\\0&-8&8\\0&6&-5\end{bmatrix}

Step 5: R2 = R2/-8 (to make A[2][2] = 1)

\begin{bmatrix}1&3&-1\\0&1&-1\\0&6&-5\end{bmatrix}



Step 6: R1 = R1 – R2 (to make A[1][3] = 0)

\begin{bmatrix}1&2&0\\0&1&-1\\0&6&-5\end{bmatrix}

Step 7: R3 – 6R2 (to make A[3][2] = 0)

\begin{bmatrix}1&2&0\\0&1&-1\\0&0&1\end{bmatrix}

Step 8: R2 = R2 + R3 (to make A[2][3] = 0)

\begin{bmatrix}1&2&0\\0&1&0\\0&0&1\end{bmatrix}

Step 9: R1 = R1 – 2R2 (to make A[1][2] = 0)

\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}

Now perform the same operation as above on the identity matrix. Result after each similar operations as above on the identity matrix, we get:



Identity \ matrix(I) = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}

Step 1: Interchange R2 and R3 rows

\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}

Step 2: R1 = R1 + R3

\begin{bmatrix} 1 & 1 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}

Step 3: R2 = R2 – 3R3

\begin{bmatrix} 1 & 1 & 0\\ 0 & -3 & 1\\ 0 & 1 & 0 \end{bmatrix}

Step 4: R3 = R3 + R1

\begin{bmatrix} 1 & 1 & 0\\ 0 & -3 & 1\\ 1 & 2 & 0 \end{bmatrix}

Step 5: R2 = R2/-8



\begin{bmatrix} 1 & 1 & 0\\ 0 & 3/8 & -1/8\\ 1 & 2 & 0 \end{bmatrix}

Step 6: R1 = R1 – R2

\begin{bmatrix} 1 & 5/8 & 1/8\\ 0 & 3/8 & -1/8\\ 1 & 2 & 0 \end{bmatrix}

Step 7: R3 – 6R2

\begin{bmatrix} 1 & 5/8 & 1/8\\ 0 & 3/8 & -1/8\\ 1 & -1/4 & 3/4 \end{bmatrix}

Step 8: R2 = R2 + R3

\begin{bmatrix} 1 & 5/8 & 1/8\\ 0 & 1/8 & 5/8\\ 1 & -1/4 & 3/4 \end{bmatrix}

Step 9: R1 = R1 – 2R2

\begin{bmatrix} -1 & 3/8 & -9/8\\ 1 & 1/8 & 5/8\\ 1 & -1/4 & 3/4 \end{bmatrix}

So, the inverse of matrix A is:



A^{-1}  = \begin{bmatrix} -1 & 3/8 & -9/8\\ 1 & 1/8 & 5/8\\ 1 & -1/4 & 3/4 \end{bmatrix}

Example 2: Find the inverse of the following matrix by elementary operations?

A = \begin{bmatrix} 1& 1 & 0\\ 0 & -1 & 0\\ 0 & 0 & 1 \end{bmatrix}

Solution:

Step 1: R1 = R1 + R2

\begin{bmatrix} 1& 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 1 \end{bmatrix}

Step 2: R2 = R2 x -1

\begin{bmatrix} 1& 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}

Similar operations on the identity matrix will result in:

A^{-1} = \begin{bmatrix} 1& 1 & 0\\ 0 & -1 & 0\\ 0 & 0 & 1 \end{bmatrix}

Example 3: Find the inverse of the following matrix by elementary operations?

A = \begin{bmatrix} 1& 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 1 \end{bmatrix}



Solution:

Step 1: Swap R2 and R3

\begin{bmatrix} 1& 0 & 0\\ 0 & 1& 1\\ 0 & 0 & 1 \end{bmatrix}

Step 2: R2 = R2 – R3

\begin{bmatrix} 1& 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}

Similar operations on the identity matrix will result in:

A^{-1} = \begin{bmatrix} 1& 0 & 0\\ 0 & -1 & 1\\ 0 & 1 & 0 \end{bmatrix}




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